Answer
$\dfrac{7 \pi}{6} + 2k\pi, \dfrac{11 \pi }{6}+ 2k \pi$
Work Step by Step
$2 \cos{x} - \dfrac{1}{\cos{x}} + \dfrac{\sin{x}}{\cos{x}} = 0$
$2 \cos{x} = 1 + \sin{x} =0$
$2- 2 \sin^2{x} - 1 + \sin{x}=0$
$2 \sin^2{x} = \sin{x} - 1 = 0$
Using the quadratic formula:
$\sin{x} = \dfrac{1 \pm \sqrt{1-4(2)(-1)}}{2(2)} = \dfrac{1\pm 3}{4}$
$ \sin{x} = 1 \hspace{30pt} \sin{x} = -\dfrac{1}{2}$
$x = \dfrac{7 \pi}{6} + 2k\pi, \dfrac{11 \pi }{6}+ 2k \pi$
